For the reaction CH4 + 2 O2 → CO2 + 2 H2O(l), calculate ΔHrxn using ΔHf° values: ΔHf°(CO2) = -393.5 kJ/mol, ΔHf°(H2O,l) = -285.8 kJ/mol, ΔHf°(CH4) = -74.8 kJ/mol.

• A. -890 kJ
• B. -393.5 kJ
• C. -1000 kJ
• D. -280 kJ

Answer: (A)

The calculation tests your ability to use standard enthalpies of formation to find the enthalpy change of a reaction. Use ΔHrxn = Σ ΔHf°(products) − Σ ΔHf°(reactants), taking into account stoichiometric coefficients.

For the products: one mole of CO2 with ΔHf° = -393.5 kJ/mol and two moles of H2O(l) with ΔHf° = -285.8 kJ/mol.

Sum of products = (-393.5) + 2(-285.8) = -393.5 - 571.6 = -965.1 kJ.

For the reactants: one mole of CH4 with ΔHf° = -74.8 kJ/mol and 2 moles of O2, for which ΔHf° = 0.

Sum of reactants = (-74.8) + 2(0) = -74.8 kJ.

ΔHrxn = -965.1 − (−74.8) = -965.1 + 74.8 = -890.3 kJ ≈ -890 kJ.

So the reaction releases about 890 kJ of energy (exothermic). The large negative value comes from forming CO2 and especially liquid water, which have substantial negative formation energies, outweighing the methane’s contribution.